Absorption and stimulated emission of radiation

Let us use some of the results of time-dependent perturbation theory to investigate the interaction of an atomic electron with classical (i.e., non-quantized) electromagnetic radiation.

The unperturbed Hamiltonian is

$$H_0 = \frac{p^2}{2 m_e} + V_0(r).$$ (824)

The standard classical prescription for obtaining the Hamiltonian of a particle of charge $q$ in the presence of an electromagnetic field is

$${\bf p} \rightarrow {\bf p} + q { \bf A},$$ (825)

$$H \rightarrow H - q \phi,$$ (826)

where ${\bf A}(\bf r)$ is the vector potential and $\phi({\bf r})$ is the scalar potential. Note that

$${\bf E} = - \nabla\phi - \frac{\partial {\bf A}}{\partial t},$$ (827)

$${\bf B} = \nabla\times {\bf A}.$$ (828)

This prescription also works in quantum mechanics. Thus, the Hamiltonian of an atomic electron placed in an electromagnetic field is

$$H = \frac{\left({\bf p} - e {\bf A}\right)^2 }{2 m_e}+ e \phi + V_0(r),$$ (829)

where ${\bf A}$ and $\phi$ are functions of the position operators. The above equation can be written

$$H = \frac{ \left(p^2 -e {\bf A}\!\cdot \! {\bf p} -e {\b... ...\!\cdot\!{\bf A} + e^2 A^2\right)}{2 m_e}+ e \phi + V_0(r).$$ (830)

Now,

$${\bf p}\!\cdot\!{\bf A} = {\bf A}\!\cdot \! {\bf p},$$ (831)

provided that we adopt the gauge $\nabla\!\cdot\!{\bf A} = 0$. Hence,

$$H = \frac{p^2}{2 m_e} -\frac{e {\bf A}\!\cdot\!{\bf p}}{m_e} +\frac{ e^2 A^2}{2 m_e}+ e \phi + V_0(r).$$ (832)

Suppose that the perturbation corresponds to a monochromatic plane-wave, for which

$$\phi = 0,$$ (833)

$${\bf A} = 2 A_0 \mbox{\boldmath$\epsilon$} \cos\!\left (\frac{\omega}{c} {\bf n}\!\cdot\!{\bf r} - \omega t\right),$$ (834)

where $\mbox{\boldmath$\epsilon$}$ and ${\bf n}$ are unit vectors which specify the direction of polarization and the direction of propagation, respectively. Note that $\mbox{\boldmath$\epsilon$}\!\cdot\!{\bf n} = 0$. The Hamiltonian becomes

$$H = H_0 + H_1(t),$$ (835)

with

$$H_0 = \frac{p^2}{2 m_e} + V(r),$$ (836)

and

$$H_1 \simeq -\frac{e {\bf A}\!\cdot\!{\bf p}}{m_e},$$ (837)

where the $A^2$ term, which is second order in $A_0$, has been neglected.

The perturbing Hamiltonian can be written

$$H_1 = - \frac{e A_0 \mbox{\boldmath$\epsilon$}\!\cdot\!{... ...ega/c) {\bf n}\!\cdot\!{\bf r} + {\rm i} \omega t]\right).$$ (838)

This has the same form as Eq. (812), provided that

$$V = - \frac{e A_0 \mbox{\boldmath$\epsilon$}\!\cdot\!{\b... ...{m_e} \exp[-{\rm i} (\omega/c) {\bf n}\!\cdot\!{\bf r} ]$$ (839)

It is clear, by analogy with the previous analysis, that the first term on the right-hand side of Eq. (838) describes the absorption of a photon of energy $\hbar\omega$, whereas the second term describes the stimulated emission of a photon of energy $\hbar\omega$. It follows from Eq. (822) that the rate of absorption is

$$w_{i\rightarrow n} = \frac{2\pi}{\hbar} \frac{e^2}{m_e^{ 2}}... ...\bf p} \vert i\rangle\vert^2 \delta(E_n-E_i -\hbar\omega).$$ (840)

The absorption cross-section is defined as the ratio of the power absorbed by the atom to the incident power per unit area in the electromagnetic field. Now the energy density of an electromagnetic field is

$$U = \frac{1}{2}\left(\frac{\epsilon_0 E_0^{ 2}}{2}+ \frac{B_0^{ 2}}{2 \mu_0} \right),$$ (841)

where $E_0$ and $B_0=E_0/c= 2 A_0 \omega/c$ are the peak electric and magnetic field-strengths, respectively. The incident power per unit area of the electromagnetic field is

$$c U = 2 \epsilon_0 c \omega^2 \vert A_0\vert^2.$$ (842)

Now,

$$\sigma_{\rm abs} = \frac{\hbar \omega w_{i\rightarrow n}}{c U},$$ (843)

so

$$\sigma_{\rm abs} = \frac{\pi e^2}{\epsilon_0 m_e^{ 2} \... ...\bf p} \vert i\rangle\vert^2 \delta(E_n-E_i -\hbar\omega).$$ (844)